Question

A smooth inclined plane is inclined at an angle $\theta$ with the horizontal. A body starts from rest and slides down the inclined surface, then the time taken by the body to reach the bottom is:

• A. $\sqrt{\frac{2h}{g}}$
• B. $\sqrt{\frac{2l}{g}}$
• C. $\frac{1}{sin\theta }\sqrt{\frac{2h}{g}}$
• D. $sin\theta \sqrt{\frac{2h}{g}}$

Answer 1

The correct option is C $\frac{1}{sin\theta }\sqrt{\frac{2h}{g}}$


$ma=mgsin\theta a=gsin\theta S=ut+\frac{1}{2}a{t}^{2}u=0\left(startsfromrest\right)l=0+\frac{1}{2}gsin\theta t^{2}t=\sqrt{\frac{2l}{gsin\theta }}[\frac{h}{l}=sin\theta ]t=\sqrt{\frac{2}{gsin\theta }-\frac{h}{sin\theta }}t=\frac{1}{sin\theta }\sqrt{\frac{2h}{g}}$